Lecture 17:Inference Michael Fourman - PowerPoint PPT Presentation

Lecture 17:Inference Michael Fourman

https://www.youtube.com/watch?v=Lvcnx6-0GhA An argument is a connected series of statements 
 to establish a proposition. 2

Is this a valid argument? • Assumptions: If the races are fixed or the gambling houses are crooked, then the tourist trade will decline. If the tourist trade declines then the police force will be happy. The police force is never happy. • Conclusion: The races are not fixed The argument is valid iff 
 2 if the assumptions are all true 
 then the conclusion is true 3

RF the Races are Fixed GC the Gambling houses are Crooked TT the Tourist Trade will decline PH the Police force will be Happy  • If the races are fixed or the gambling houses  are crooked then the tourist trade will decline.      (RF ∨ GC) → TT    Assumptions: • If the tourist trade declines then the police   force will be happy. TT → PH       • The police force is never happy. ¬ PH  Conclusion: • The races are not fixed. ¬ RF The argument is valid i ff the following entailment is valid: (RF ∨ GC) → TT , TT → PH , ¬ PH | = ¬ RF 4

We could check the validity of the entailment by checking all sixteen assignments of truth values to the four basic propositions. Can we do do less work? Consider our example (RF ∨ GC) → TT , TT → PH , ¬ PH | = ¬ RF Remember that an entailment is valid 
 unless there is a counterexample. A counterexample is an assignment of truth values that makes everything on the left true, 
 and everything on the right false. 5

A counterexample is an assignment of truth values that makes everything on the left true, 
 and everything on the right false. The basic idea: for each entailment Γ ⊨ Δ show that if there is a counterexample to this entailment then there is a counterexample to some simpler entailment. Consider: (RF ∨ GC) → TT , TT → PH , ¬ PH | = ¬ RF (1) (RF ∨ GC) → TT , TT → PH , RF | = PH (2) Any counterexample to (1) is a counterexample to (2) (and vice versa ). 6

Consider: (RF ∨ GC) → TT , TT → PH , ¬ PH | = ¬ RF (1) (RF ∨ GC) → TT , TT → PH , RF | = PH (2) Any counterexample to (1) is a counterexample to (2) (and vice versa ). (2) is simpler - there are fewer logical operators If (2) is valid, there is no counterexample, so (1) is also valid We write this as a rule (RF ∨ GC) → TT , TT → PH , RF | = PH (RF ∨ GC) → TT , TT → PH , ¬ PH | = ¬ RF 7

There is a counterexample to the conclusion, 
 iff there is a counterexample to the assumption. Therefore ∴ If the assumption of the rule (above the line) is valid, then the conclusion (below the line) is valid. (RF ∨ GC) → TT , TT → PH , RF | = PH (RF ∨ GC) → TT , TT → PH , ¬ PH | = ¬ RF 8

Now consider (RF ∨ GC) → TT , TT → PH , RF | = PH (2) Any counterexample will make TT → PH true so it will either make TT false, in which case it is a counterexample to (RF ∨ GC) → TT , RF | = PH , TT (3) or make PH true, in which case it is a counterexample to (RF ∨ GC) → TT , PH , RF | = PH (4) (or both). There is a counter-example to (2) i ff there is a counter- example to (at least) one of (3), (4). 9

Now consider (RF ∨ GC) → TT , TT → PH , RF | = PH (2) Any counterexample will make TT → PH true so it will either make TT false, in which case it is a counterexample to (RF ∨ GC) → TT , RF | = PH , TT (3) or make PH true, in which case it is a counterexample to (RF ∨ GC) → TT , PH , RF | = PH (4) This gives a rule: (RF ∨ GC) → TT , RF | = PH , TT (RF ∨ GC) → TT , PH , RF | = PH (RF ∨ GC) → TT , TT → PH , RF | = PH There is a counter-example to the conclusion i ff there is a counter-example to (at least) one of the assumptions. 10

Putting these two rules together we start to build a proof tree (RF ∨ GC) → TT , RF | = PH , TT (RF ∨ GC) → TT , PH , RF | = PH (RF ∨ GC) → TT , TT → PH , RF | = PH (RF ∨ GC) → TT , TT → PH , ¬ PH | = ¬ RF If we have a counterexample to the conclusion then we have a counterexample to at least one of the assumptions 11

Now consider (RF ∨ GC) → TT , PH , RF | = PH (4) Any counterexample would make PH true and make PH false, but this is impossible, so there are no counterexamples. We draw a line over (4) to make a rule with no assumptions. (RF ∨ GC) → TT , PH , RF | = PH We still have the key property: • there is a counterexample to the conclusion i ff there is a counterexample to (at least) one of the assumptions 12

Only one assumption remains (RF ∨ GC) → TT , RF | = PH , TT (RF ∨ GC) → TT , PH , RF | = PH (RF ∨ GC) → TT , TT → PH , RF | = PH (RF ∨ GC) → TT , TT → PH , ¬ PH | = ¬ RF If we have a counterexample to the conclusion then we have a counterexample to at least one of the assumptions. Our next step should be familiar. 13

We follow a pattern used earlier Γ ` A, ∆ Γ , B ` ∆ ( ! L ) Γ , A ! B ` ∆ ✓ ◆ with Γ = RF , A = RF ∨ GC , B = TT , ∆ = PH, TT RF | = PH , TT , RF ∨ GC TT , RF | = PH , TT (RF ∨ GC) → TT , RF | = PH , TT (RF ∨ GC) → TT , PH , RF | = PH (RF ∨ GC) → TT , TT → PH , RF | = PH (RF ∨ GC) → TT , TT → PH , ¬ PH | = ¬ RF 14

Another pattern we used earlier Γ , A ` ∆ , A ( I ) ( with A = TT , Γ = RF , ∆ = PH ) RF | = PH , TT , RF ∨ GC TT , RF | = PH , TT (RF ∨ GC) → TT , RF | = PH , TT (RF ∨ GC) → TT , PH , RF | = PH (RF ∨ GC) → TT , TT → PH , RF | = PH (RF ∨ GC) → TT , TT → PH , ¬ PH | = ¬ RF 15

Now consider RF | = PH , TT , RF ∨ GC Any counterexample will make both RF and GC false, so it is a counterexample to RF | = PH , TT , RF , GC This gives a rule RF | = PH , TT , RF , GC RF | = PH , TT , RF ∨ GC A valuation is a counter-example to the conclusion i ff it is a counter-example to the assumption. 16

The pattern for this rule is Γ ` A, B, ∆ Γ ` A _ B, ∆ ( _ R ) Our proof is almost done RF | = PH , TT , RF , GC RF | = PH , TT , RF ∨ GC TT , RF | = PH , TT (RF ∨ GC) → TT , RF | = PH , TT (RF ∨ GC) → TT , PH , RF | = PH (RF ∨ GC) → TT , TT → PH , RF | = PH (RF ∨ GC) → TT , TT → PH , ¬ PH | = ¬ RF 17

Immediate! Γ , A ` ∆ , A ( I ) RF | = PH , TT , RF , GC RF | = PH , TT , RF ∨ GC TT , RF | = PH , TT (RF ∨ GC) → TT , RF | = PH , TT (RF ∨ GC) → TT , PH , RF | = PH (RF ∨ GC) → TT , TT → PH , RF | = PH (RF ∨ GC) → TT , TT → PH , ¬ PH | = ¬ RF 18

Gentzen’s Rules (I) Γ , A ` ∆ , A ( I ) 1924 1945 Γ , A, B ` ∆ Γ ` A, B, ∆ Γ , A ^ B ` ∆ ( ^ L ) Γ ` A _ B, ∆ ( _ R ) Γ , A ` ∆ Γ , B ` ∆ Γ ` A, ∆ Γ ` B, ∆ ( _ L ) ( ^ R ) Γ , A _ B ` ∆ Γ ` A ^ B, ∆ a sequent, Γ ⊢ Δ 
 where Γ and Δ are finite sets of expressions is valid iff 
 whenever every expression in Γ is true some expression in Δ is true Gerhard Karl Erich Gentzen (November 24, 1909 – August 4, 1945) 19

Gentzen’s Rules (I) Γ , A ` ∆ , A ( I ) Γ , A, B ` ∆ Γ ` A, B, ∆ Γ , A ^ B ` ∆ ( ^ L ) Γ ` A _ B, ∆ ( _ R ) Γ , A ` ∆ Γ , B ` ∆ Γ ` A, ∆ Γ ` B, ∆ ( _ L ) ( ^ R ) Γ , A _ B ` ∆ Γ ` A ^ B, ∆ a counterexample to the sequent Γ ⊢ Δ , 
 is a valuation that makes every expression in Γ true and every expression in Δ false (a sequent is valid iff it has no counterexample) 20

A, B ` A, B ( I ) A ^ B ` A, B ( ^ L ) A ^ B ` A _ B ( _ R ) 21

A rule Γ , A ` B, ∆ Γ ` A ! B, ∆ ( ! R ) A valuation is a counterexample to the top line iff it is a counterexample to the bottom line 22

Another rule Γ ` A, ∆ Γ , B ` ∆ ( ! L ) Γ , A ! B ` ∆ A valuation is a counterexample to the bottom line iff it is a counterexample to at least one of the entailments on the top line 23

a valuation is a it is a counterexample iff counterexample to to at least one the conclusion assumption Γ , A ` ∆ , A ( I ) Γ , A, B ` ∆ Γ ` A, B, ∆ Γ , A ^ B ` ∆ ( ^ L ) Γ ` A _ B, ∆ ( _ R ) Γ , A ` ∆ Γ , B ` ∆ Γ ` A, ∆ Γ ` B, ∆ ( _ L ) ( ^ R ) Γ , A _ B ` ∆ Γ ` A ^ B, ∆ Γ ` A, ∆ Γ , B ` ∆ Γ , A ` B, ∆ ( ! L ) Γ ` A ! B, ∆ ( ! R ) Γ , A ! B ` ∆ Γ ` A, ∆ Γ , A ` ∆ Γ , ¬ A ` ∆ ( ¬ L ) Γ ` ¬ A, ∆ ( ¬ R ) 24

a valuation is a it is a counterexample iff counterexample to to at least one the conclusion assumption Γ , A ` ∆ , A ( I ) Γ , A, B ` ∆ Γ ` A, B, ∆ Γ , A ^ B ` ∆ ( ^ L ) Γ ` A _ B, ∆ ( _ R ) Γ , A ` ∆ Γ , B ` ∆ Γ ` A, ∆ Γ ` B, ∆ ( _ L ) ( ^ R ) Γ , A _ B ` ∆ Γ ` A ^ B, ∆ Γ ` A, ∆ Γ , B ` ∆ Γ , A ` B, ∆ ( ! L ) Γ ` A ! B, ∆ ( ! R ) Γ , A ! B ` ∆ Γ ` A, ∆ Γ , A ` ∆ Γ , ¬ A ` ∆ ( ¬ L ) Γ ` ¬ A, ∆ ( ¬ R ) 25

Γ , A ` ∆ , A ( I ) Γ ` A, ∆ Γ , B ` ∆ Γ , A ` B, ∆ ( ! L ) Γ ` A ! B, ∆ ( ! R ) Γ , A ! B ` ∆ ?? A ! ( B ! C ) ` B ! ( A ! C ) 26

Γ , A ` ∆ , A ( I ) Γ ` A, ∆ Γ , B ` ∆ Γ , A ` B, ∆ ( ! L ) Γ ` A ! B, ∆ ( ! R ) Γ , A ! B ` ∆ ?? this goal A ! ( B ! C ) ` B ! ( A ! C ) Γ , A → B ⊢ Δ matches the conclusion of ( → L ) where Γ is empty Δ is B → ( A → C ) A is A B is B → C 27

Γ , A ` ∆ , A ( I ) Γ ` A, ∆ Γ , B ` ∆ Γ , A ` B, ∆ ( ! L ) Γ ` A ! B, ∆ ( ! R ) Γ , A ! B ` ∆ ?? this goal : A ! ( B ! C ) ` B ! ( A ! C ) matches Γ ⊢ Α → B , Δ which is the conclusion of ( → R ) where Γ is Α → ( B → C ) Δ is empty A is B B is Α → C 28